chapter six thermochemistry · thermochemistry questions 9. a coff ee-cup calor imeter is at cons...

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CHAPTER SIX

THERMOCHEMISTRY

Questions

9. A coffee-cup calorimeter is at constant (atmospheric) pressure. The heat released or gained at constantpressure is )H. A bomb calorimeter is at constant volume. The heat released or gained at constantvolume is )E.

10. Plot a represents an exothermic reaction. In an exothermic process, the bonds in the product moleculesare stronger (on average) than those in the reactant molecules. The net result is that the quantity ofenergy )(PE) is transferred to the surroundings as heat when reactants are converted to products.

For an endothermic process, energy flows into the system as heat to increase the potential energy ofthe system. In an endothermic process, the products have higher potential energy (weaker bonds onaverage) than the reactants.

11. The specific heat capacities are: 0.89 J/gC°C (Al) and 0.45 J/gC°C (Fe)Al would be the better choice. It has a higher heat capacity and a lower density than Fe. Using Al,the same amount of heat could be dissipated by a smaller mass, keeping the mass of the amplifierdown.

2 212. H O(l) ö H O(g); Heat must be added to boil water so q is positive. Since a certain quantity of

2 2H O(g) occupies a much larger volume than the same quantity of H O(l), an expansion will occur as

2 2H O(l) is converted to H O(g). Therefore, w will be negative for this process, i.e., the system doeswork on the surroundings when the expansion occurs.

13. A state function is a function whose change depends only on the initial and final states and not on howone got from the initial to the final state. If H and E were not state functions, the law of conservationof energy (first law) would not be true.

14. If Hess's law were not true it would be possible to create energy by reversing a reaction using adifferent series of steps. This violates the law of conservation of energy (first law). Thus, Hess's lawis another statement of the law of conservation of energy.

15. In order to compare values of )H to each other, a common reference (or zero) point must be chosen.The definition of establishes the pure elements in their standard states as that common referencepoint.

216. Advantages: H burns cleanly (less pollution) and gives a lot of energy per gram of fuel.

CHAPTER 6 THERMOCHEMISTRY130

Disadvantages: Expense and storage.

Exercises

Potential and Kinetic Energy

17. KE = mv ; Convert mass and velocity to SI units. 1 J = 2

Mass = 5.25 oz × = 0.149 kg

Velocity =

KE = mv = × 0.149 kg × = 150 J2

18. KE = mv = × = 2.0 × 10 J2 -2

19. KE = mv = × 2.0 kg × = 1.0 J; KE = mv = × 1.0 kg × = 2.0 J2 2

The 1.0 kg object with a velocity of 2.0 m/s has the greater kinetic energy.

20. Ball A: PE = mgz = 2.00 kg × × 10.0 m = = 196 J

At Point I: All of this energy is transferred to Ball B. All of B's energy is kinetic energy at this

totalpoint. E = KE = 196 J. At point II, the sum of the total energy will equal 196 J.

At Point II: PE = mgz = 4.00 kg × × 3.00 m = 118 J

totalKE = E - PE = 196 J - 118 J = 78 J

Heat and Work

21. a. )E = q + w = 51 kJ + (-15 kJ) = 36 kJ

b. )E = 100. kJ + (-65 kJ) = 35 kJ c. )E = -65 + (-20.) = -85 kJ

d. When the system delivers work to the surroundings, w < 0. This is the case in all these examples,a, b and c.

22. a. )E = q + w = -47 kJ + 88 kJ = 41 kJ

CHAPTER 6 THERMOCHEMISTRY 131

b. )E = 82 + 47 = 129 kJ c. )E = 47 + 0 = 47 kJ

d. When the surroundings deliver work to the system, w > 0. This is the case for a and b.

23. )E = q + w = -125 + 104 = -21 kJ

1 224. Step 1: )E = q + w = 72 J + 35 J = 107 J; Step 2: )E = 35 J - 72 J = -37 J

overall 1 2)E = )E + )E = 107 J - 37 J = 70. J

f i25. w = -P)V = -P × (V - V ) = -2.0 atm × (5.0 × 10 L - 5.0 L) = -2.0 atm × (-5.0 L) = 10. L atm-3

We can also calculate the work in Joules.

1 atm = 1.013 × 10 Pa = 1.013 × 10 ; 1 L = 1000 cm = 1 × 10 m5 5 3 -3 3

1 L atm = 1 × 10 m × 1.013 × 10 = 101.3 = 101.3 J-3 3 5

w = 10. L atm × = 1013 J = 1.0 × 10 J3

26. In this problem q = w = -950. J

-950. J × = -9.38 L atm of work done by the gases.

f f fw = -P)V, -9.38 L atm = atm × (V - 0.040 L), V - 0.040 = 11.0 L, V = 11.0 L

27. q = molar heat capacity × mol × )T = × 39.1 mol × (38.0 - 0.0) °C = 30,900 J = 30.9 kJ

w = -P)V = -1.00 atm × (998 L - 876 L) = -122 L atm × = -12,400 J = -12.4 kJ

)E = q + w = 30.9 kJ + (-12.4 kJ) = 18.5 kJ

2 228. H O(g) ÷ H O(l); )E = q + w; q = -40.66 kJ; w = -P)V

2 2Volume of 1 mol H O(l) = 1 mol H O(l) × = 18.1 cm = 18.1 mL3

w = -P)V = - 1.00 atm × (0.0181 L - 30.6 L) = 30.6 L atm × = 3.10 × 10 J = 3.10 kJ3

)E = q + w = -40.66 kJ + 3.10 kJ = -37.56 kJ

Properties of Enthalpy

29. This is an endothermic reaction so heat must be absorbed in order to convert reactants into products.


The high temperature environment of internal combustion engines provides the heat.

30. One should try to cool the reaction mixture or provide some means of removing heat since the reaction

2 4is very exothermic (heat is released). The H SO (aq) will get very hot and possibly boil unless coolingis provided.

31. a. Heat is absorbed from the water (it gets colder) as KBr dissolves, so this is an endothermic process.

4b. Heat is released as CH is burned, so this is an exothermic process.

2 4c. Heat is released to the water (it gets hot) as H SO is added, so this is an exothermicprocess.

d. Heat must be added (absorbed) to boil water, so this is an endothermic process.

32. a. The combustion of gasoline releases heat, so this is an exothermic process.

2 2b. H O(g) ÷ H O(l); Heat is released when water vaper condenses, so this is an exothermicprocess.

c. To convert a solid to a gas, heat must be absorbed, so this is an endothermic process.

d. Heat must be added (absorbed) in order to break a bond, so this is an endothermic process.

2 2 333. 4 Fe(s) + 3 O (g) v 2 Fe O (s) )H = -1652 kJ; Note that 1652 kJ of heat are released when 4

2 2 3mol Fe react with 3 mol O to produce 2 mol Fe O .

a. 4.00 mol Fe × = -1650 kJ heat released

2 3b. 1.00 ml Fe O × = -826 kJ heat released

c. 1.00 g Fe × = -7.39 kJ heat released

2 2d. 10.0 g Fe × = 0.179 mol Fe; 2.00 g O × = 0.0625 mol O

2 20.179 mol Fe/0.0625 mol O = 2.86; The balanced equation requires a 4 mol Fe/3 mol O = 1.33

2 2mol ratio. O is limiting since the actual mol Fe/mol O ratio is less than the required mol ratio.

20.0625 mol O × = -34.4 kJ heat released

234. a. 1.00 mol H O × = -286 kJ heat released


2b. 4.03 g H × = -572 kJ heat released

2c. 186 g O × = -3320 kJ heat released

2d. = = 8.2 × 10 mol H6

28.2 × 10 mol H × = -2.3 × 10 kJ heat released6 9

35. From Sample Exercise 6.3, q = 1.3 × 10 J. Since the heat transfer process is only 60.%8

efficient, the total energy required is: 1.3 × 10 J × = 2.2 × 10 J8 8

3 8 3 8 mass C H = 2.2 × 10 J × = 4.4 × 10 g C H8 3

436. a. 1.00 g CH × = -55.5 kJ

b. PV = nRT, n = = 39.8 mol

39.8 mol × = -3.55 × 10 kJ4

Calorimetry and Heat Capacity

37. Specific heat capacity is defined as the amount of heat necessary to raise the temperature of one gram

2of substance by one degree Celsius. Therefore, H O(l) with the largest heat capacity value requires

2the largest amount of heat for this process. The amount of heat for H O(l) is:

energy = s × m × )T = × 25.0 g × (37.0°C - 15.0°C) = 2.30 × 10 J3

The largest temperature change when a certain amount of energy is added to a certain mass ofsubstance will occur for the substance with the smallest specific heat capacity. This is Hg(l), and thetemperature change for this process is:

)T = = 140°C


38. a. s = specific heat capacity = since )T(K) = )T(°C).

energy = s × m × )T = × 150.0 g × (298 K - 273 K) = 9.0 × 10 J2

b. molar heat capacity = =

c. 1250 J = × m × (15.2°C - 12.0°C), m = = 1.6 × 10 g Ag3

39. The units for specific heat capacity (s) are J/gC°C. s =

Molar heat capacity =

40. s = = 0.139 J/gC°C

Molar heat capacity =

41. * Heat loss by hot water * = * Heat gain by cooler water *

The magnitude of heat loss and heat gain are equal in calorimetry problems. The only difference isthe sign (positive or negative). To avoid sign errors, keep all quantities positive and, if necessary,deduce the correct signs at the end of the problem. Water has a specific heat capacity = s = 4.18J/°CCg = 4.18 J/KCg ()T in °C = )T in K).

fHeat loss by hot water = s × m × )T = × 50.0 g × (330. K - T )

fHeat gain by cooler water = × 30.0 g × (T - 280. K); Heat loss = Heat gain, so:

f f f f × (330. K - T ) = × (T - 280. K), 6.90 × 10 - 209 T = 125 T - 3.50 × 104 4

f f334 T = 1.040 × 10 , T = 311 K5

Note that the final temperature is closer to the temperature of the more massive hot water, which is asit should be.

42. Heat loss by Al + heat loss by Fe = heat gain by water; Keeping all quantities positive to avoid signerror:

f f × 5.00 g Al × (100.0°C - T ) + × 10.00 g Fe × (100.0 - T )


2 f = × 97.3 g H O × (T - 22.0°C)

f f f f f f4.5(100.0 - T ) + 4.5(100.0 - T ) = 407(T - 22.0), 450 - 4.5 T + 450 - 4.5 T = 407 T - 8950

f f416 T = 9850, T = 23.7°C

43. Heat gained by water = heat loss by metal = s × m × )T where s = specific heat capacity.

Heat gain = × 150.0 g × (18.3°C - 15.0°C) = 2100 J

A common error in calorimetry problems is sign errors. Keeping all quantities positive helps eliminate sign errors.

heat loss = 2100 J = s × 150.0 g × (75.0°C - 18.3°C), s = = 0.25 J/gC°C

44. Heat gain by water = heat loss by Cu; Keeping all quantities positive to avoid sign errors:

× m × (24.9°C - 22.3°C) = × 110. g Cu × (82.4°C - 24.9°C)

211 m = 1300, m = 120 g H O

345. 50.0 × 10 L × 0.100 mol/L = 5.00 × 10 mol of both AgNO and HCl are reacted. Thus,-3 -3

5.00 × 10 mol of AgCl will be produced since there is a 1:1 mol ratio between reactants.-3

Heat lost by chemicals = Heat gained by solution

Heat gain = × 100.0 g × (23.40 - 22.60)°C = 330 J

Heat loss = 330 J; This is the heat evolved (exothermic reaction) when 5.00 × 10 mol of AgCl is-3

produced. So q = -330 J and )H (heat per mol AgCl formed) is negative with a value of:

)H = = -66 kJ/mol

Note: Sign errors are common with calorimetry problems. However, the correct sign for )Hcan easily be determined from the )T data, i.e., if )T of the solution increases, then the reactionis exothermic since heat was released, and if )T of the solution decreases, then the reaction isendo- thermic since the reaction absorbed heat from the water. For calorimetry problems, keepall quantities positive until the end of the calculation, then decide the sign for )H. This will helpeliminate sign errors.

4 3 4 346. NH NO (s) ÷ NH (aq) + NO (aq) )H = ?; mass of solution = 75.0 g + 1.60 g = 76.6 g+ -

4 3Heat lost by solution = Heat gained as NH NO dissolves. To help eliminate sign errors, we willkeep all quantities positive (q and )T), then deduce the correct sign for )H at the end of the

4 3 4 3problem. Here, since temperature decreases as NH NO dissolves, heat is absorbed as NH NO


dissolves, so it is an endothermic process ()H is positive).

4 3Heat lost by solution = × 76.6 g × (25.00 - 23.34)°C = 532 J = heat gained as NH NO dissolves

4 3)H = = 26.6 kJ/mol NH NO dissolving

247. Since )H is exothermic, the temperature of the solution will increase as CaCl (s) dissolves.Keeping all quantities positive:

2 2Heat loss as CaCl dissolves = 11.0 g CaCl × = 8.08 kJ

fHeat gained by solution = 8.08 × 10 J = × (125 + 11.0) g × (T - 25.0°C)3

f fT - 25.0°C = = 14.2°C, T = 14.2°C + 25.0°C = 39.2°C

48. 0.100 L × = 5.00 × 10 mol HCl-2

20.300 L × = 3.00 × 10 mol Ba(OH)-2

2To react with all the HCl present, 5.00 × 10 /2 = 2.50 × 10 mol Ba(OH) are required. Since-2 -2

23.00 × 10 mol Ba(OH) are present, HCl is the limiting reactant.-2

5.00 × 10 mol HCl × = 2.95 kJ of heat is evolved by reaction.-2

Heat gained by solution = 2.95 × 10 J = × 400.0 g × )T3

f i f f)T = 1.76°C = T - T = T - 25.0°C, T = 26.8°C

49. First, we need to get the heat capacity of the calorimeter from the combustion of benzoicacid.

Heat lost by combustion = Heat gained by calorimeter

Heat loss = 0.1584 g × = 4.185 kJ

cal calHeat gain = 4.185 kJ = C × )T, C = = 1.65 kJ/°C

Now we can calculate the heat of combustion of vanillin. Heat loss = Heat gain

Heat gain by calorimeter = × 3.25°C = 5.36 kJ

Heat loss = 5.36 kJ, which is the heat evolved by the combustion of the vanillin.


comb comb)E = = -25.2 kJ/g; )E = = -3830 kJ/mol

50. Heat gain by calorimeter = × 3.2°C = 5.0 kJ = heat loss by quinone

Heat loss = 5.0 kJ, which is the heat evolved (exothermic reaction) by the combustion of 0.1964g of quinone.

comb comb)E = = -25 kJ/g; )E = = -2700 kJ/mol

Hess's Law

51. Information given:

2 2 C(s) + O (g) ÷ CO (g) )H = -393.7 kJ

2 2CO(g) + 1/2 O (g) ÷ CO (g) )H = -283.3 kJ

Using Hess’s Law:

2 2 1 2 C(s) + 2 O (g) ÷ 2 CO (g) )H = 2(-393.7 kJ)

2 2 2 2 CO (g) ÷ 2 CO(g) + O (g) )H = -2(-283.3 kJ)

2 1 22 C(s) + O (g) ÷ 2 CO(g) )H = )H + )H = -220.8 kJ

Note: The enthalpy change for a reaction that is reversed is the negative quantity of the enthalpychange for the original reaction. If the coefficients in a balanced reaction are multiplied by aninteger, the value of )H is multiplied by the same integer.

4 4 2 2 2 comb52. C H (g) + 5 O (g) ÷ 4 CO (g) + 2 H O(l) )H = -2341 kJ

4 8 2 2 2 combC H (g) + 6 O (g) ÷ 4 CO (g) + 4 H O(l) )H = -2755 kJ

2 2 2 combH (g) + 1/2 O (g) ÷ H O(l) )H = -286 kJ

2By convention, H O(l) is produced when enthalpies of combustion are given and, since per mol

2quantities are given, the combustion reaction refers to 1 mol of that quantity reacting with O (g).

Using Hess’s Law to solve:

4 4 2 2 2 1 C H (g) + 5 O (g) ÷ 4 CO (g) + 2 H O(l) )H = -2341 kJ

2 2 4 8 2 24 CO (g) + 4 H O(l) ÷ C H (g) + 6 O (g) )H = -(-2755 kJ)

2 2 2 3 2 H (g) + O (g) ÷ 2 H O(l) )H = 2(-286 kJ)

4 4 2 4 8 1 2 3 C H (g) + 2 H (g) ÷ C H (g) )H = )H + )H + )H = -158 kJ


2 2 253. 2 NO ÷ N + 2 O )H = -(67.7 kJ)

2 2 2 4 N + 2 O ÷ N O )H = 9.7 kJ

2 2 4 2 NO (g) ÷ N O (g) )H = -58.0 kJ

2 2 354. 2 N (g) + 6 H (g) ÷ 4 NH (g) )H = -4(46 kJ)

2 2 2 6 H O(g) ÷ 6 H (g) + 3 O (g) )H = -3(-484 kJ)

2 2 2 32 N (g) + 6 H O(g) ÷ 3 O (g) + 4 NH (g) )H = 1268 kJ

No, since the reaction is very endothermic (requires a lot of heat), it would not be a practical wayof making ammonia due to the high energy costs required.

3 2 255. NO + O ÷ NO + O )H = -199 kJ

2 3 3/2 O ÷ O )H = -1/2(-427 kJ)

2O ÷ 1/2 O )H = -1/2(495 kJ)

2NO(g) + O(g) ÷ NO (g) )H = -233 kJ

6 4 2 6 4 2 256. C H (OH) ÷ C H O + H )H = 177.4 kJ

2 2 2 2 H O ÷ H + O )H = -(-191.2 kJ)

2 2 2 2 H + O ÷ 2 H O(g) )H = 2(-241.8 kJ)

2 2 2 H O(g) ÷ 2 H O(l) )H = 2(-43.8 kJ)

6 4 2 2 2 6 4 2 2C H (OH) (aq) + H O (aq) ÷ C H O (aq) + 2 H O(l) )H = -202.6 kJ

3 2 5 257. 4 HNO ÷ 2 N O + 2 H O )H = -2(-76.6 kJ)

2 2 2 3 2 N + 6 O + 2 H ÷ 4 HNO )H = 4(-174.1 kJ)

2 2 2 2 H O ÷ 2 H + O )H = -2(-285.8 kJ)

2 2 2 5 2 N (g) + 5 O (g) ÷ 2 N O (g) )H = 28.4 kJ

4 10 4 258. P O ÷ P + 5 O )H = -(-2967.3 kJ)

3 2 3 10 PCl + 5 O ÷ 10 Cl PO )H = 10(-285.7 kJ)

5 3 2 6 PCl ÷ 6 PCl + 6 Cl )H = -6(-84.2 kJ)

4 2 3 P + 6 Cl ÷ 4 PCl )H = -1225.6 kJ

4 10 5 3P O (s) + 6 PCl (g) ÷ 10 Cl PO(g) )H = - 610.1 kJ

Standard Enthalpies of Formation

59. The change in enthalpy that accompanies the formation of one mole of a compound from itselements, with all substances in their standard states, is the standard enthalpy of formation fora compound. The reactions that refer to are:

2 2 2 2Na(s) + 1/2 Cl (g) ÷ NaCl(s); H (g) + 1/2 O (g) ÷ H O(l)


2 2 6 12 6 2 6 C(graphite, s) + 6 H (g) + 3 O (g) ÷ C H O (s); Pb(s) + S(rhombic, s) + 2 O (g)

4 ÷ PbSO (s)

2 3 2 2 360. a. aluminum oxide = Al O ; 2 Al(s) + 3/2 O (g) ÷ Al O (s)

2 5 2 2 2b. C H OH(l) + 3 O (g) ÷ 2 CO (g) + 3 H O(l)

2c. NaOH(aq) + HCl(aq) ÷ H O(l) + NaCl(aq)

2 2 2 3d. 2 C(graphite, s) + 3/2 H (g) + 1/2 Cl (g) ÷ C H Cl(g)

6 6 2 2 2e. C H (l) + 15/2 O (g) ÷ 6 CO (g) + 3 H O(l)

combNote: )H values assume one mol of compound combusted.

4 4f. NH Br(s) ÷ NH (aq) + Br (aq)+ -

61. In general: )H° = - and all elements in their standard state have = 0 by definition.

3 2 4 2a. The balanced equation is: 2 NH (g) + 3 O (g) + 2 CH (g) ÷ 2 HCN(g) + 6 H O(g)

2)H° = [2 mol HCN × + 6 mol H O(g) × ]

3 4 - [2 mol NH × + 2 mol CH × ]

)H° = [2(135.1) + 6(-242)] - [2(-46) + 2(-75)] = -940. kJ

3 4 2 2 4 4 3 4b. Ca (PO ) (s) + 3 H SO (l) ÷ 3 CaSO (s) + 2 H PO (l)

)H° =

-

)H° = -6833 kJ - (-6568 kJ) = -265 kJ

3 4c. NH (g) + HCl(g) ÷ NH Cl(s)

4 3)H° = [1 mol NH Cl × ] - [1 mol NH × + 1 mol HCl × ]

)H° =

)H° = -314 kJ + 138 kJ = -176 kJ


2 5 2 2 262. a. The balanced equation is: C H OH(l) + 3 O (g) ÷ 2 CO (g) + 3 H O(g)

)H° =

= -1513 kJ - (-278 kJ) = -1235 kJ

4 2 2b. SiCl (l) + 2 H O(l) ÷ SiO (s) + 4 HCl(aq)

Since HCl(aq) is H (aq) + Cl (aq), then = 0 - 167 = -167 kJ/mol.+ -

)H° =

)H° = -1579 kJ - (-1259 kJ) = -320. kJ

2 2c. MgO(s) + H O(l) ÷ Mg(OH) (s)

)H° =

)H° = -925 kJ - (-888 kJ) = -37 kJ

3 2 263. a. 4 NH (g) + 5 O (g) ÷ 4 NO(g) + 6 H O(g); )H° = -

)H° = = -908 kJ

2 22 NO(g) + O (g) ÷ 2 NO (g)

)H° = = -112 kJ

2 2 33 NO (g) + H O(l) ÷ 2 HNO (aq) + NO(g)

)H° =

- = -140. kJ

Note: All values are assumed ± 1 kJ.


3 2 2b. 12 NH (g) + 15 O (g) ÷ 12 NO(g) + 18 H O(g)

2 2 12 NO(g) + 6 O (g) ÷ 12 NO (g)

2 2 3 12 NO (g) + 4 H O(l) ÷ 8 HNO (aq) + 4 NO(g)

2 2 4 H O(g) ÷ 4 H O(l)

3 2 3 2 12 NH (g) + 21 O (g) ÷ 8 HNO (aq) + 4 NO(g) + 14 H O(g)

The overall reaction is exothermic since each step is exothermic.

2 264. 4 Na(s) + O (g) ÷ 2 Na O(s), )H° = 2 mol = -832 kJ

2 22 Na(s) + 2 H O(l) ÷ 2 NaOH(aq) + H (g)

)H° = = -368 kJ

2 22 Na(s) + CO (g) ÷ Na O(s) + CO(g)

)H° = = -133 kJ

In both cases, sodium metal reacts with the "extinguishing agent." Both reactions are exothermic and

2each reaction produces a flammable gas, H and CO, respectively.

4 4 2 3 3 265. 3 Al(s) + 3 NH ClO (s) ÷ Al O (s) + AlCl (s) + 3 NO(g) + 6 H O(g)

)H° =

- = -2677 kJ

2 4 2 3 3 2 2 266. 5 N O (l) + 4 N H CH (l) ÷ 12 H O(g) + 9 N (g) + 4 CO (g)

)H° =

- = -4594 kJ

3 3 2 267. 2 ClF (g) + 2 NH (g) ÷ N (g) + 6 HF(g) + Cl (g) )H° = -1196 kJ

)H° = [6 ] - [2 + 2 ]


-1196 kJ = 6 mol - 2 - 2 mol

-1196 kJ = -1626 kJ - 2 + 92 kJ, =

2 4 2 2 268. C H (g) + 3 O (g) ÷ 2 CO (g) + 2 H O(l) )H° = -1411.1 kJ

)H° = -1411.1 kJ = 2(-393.5) kJ + 2(-285.8) kJ -

-1411.1 kJ = -1358.6 kJ - , = 52.5 kJ/mol

Energy Consumption and Sources

2 5 2 2 269. C H OH(l) + 3 O (g) ÷ 2 CO (g) + 3 H O(l)

)H° = [2 (-393.5 kJ) + 3(-286 kJ)] - (-278 kJ) = -1367 kJ/mol ethanol

= -29.67 kJ/g

2 370. CO(g) + 2 H (g) ÷ CH OH(l), )H° = -239 kJ - (-110.5 kJ) = -129 kJ

3 8 2 2 271. C H (g) + 5 O (g) ÷ 3 CO (g) + 4 H O(l)

3 8)H° = [3(-393.5 kJ) + 4(-286 kJ)] - [-104 kJ] = -2221 kJ/mol C H

vs. -47.7 kJ/g for octane (Sample Exercise 6.11)

The fuel values are very close. An advantage of propane is that it burns more cleanly. The boilingpoint of propane is -42°C. Thus, it is more difficult to store propane and there are extra safety hazardsassociated with using high pressure compressed gas tanks.

2 2 4 1072. Since 1 mol of C H (g) and 1 mol of C H (g) have equivalent volumes at the same T and P, then:

= 0.452


Almost twice the volume of acetylene is needed to furnish the same energy as a given volume ofbutane.

73. The molar volume of a gas at STP is 22.42 L (from Chapter 5).

44.19 × 10 kJ × = 1.05 × 10 L CH6 5

2 274. Mass of H O = 1.00 gal × = 3790 g H O

Energy required (theoretical) = s × m × )T = × 3790 g × 10.0 °C = 1.58 × 10 J5

For an actual (80.0% efficient) process, more than this quantity of energy is needed since heat isalways lost in any transfer of energy. The energy required is:

1.58 × 10 J × = 1.98 × 10 J5 5

2 2 2 2Mass of C H = 1.98 × 10 J × = 3.97 g C H5

Additional Exercises

2 2 375. a. 2 SO (g) + O (g) 6 2 SO (g) (w = -P)V); Since the volume of the piston apparatus decreasedas reactants were converted to products, w is positive (w > 0).

2 2b. COCl (g) 6 CO(g) + Cl (g); Since the volume increased, w is negative (w < 0).

2 2c. N (g) + O (g) 6 2 NO(g); Since the volume did not change, no PV work is done (w = 0).

In order to predict the sign of w for a reaction, compare the coefficients of all the product gases in thebalanced equation to the coefficients of all the reactant gases. When a balanced reaction has more molof product gases than mol of reactant gases (as in b), the reaction will expand in volume ()V positive),and the system does work on the surroundings. When a balanced reaction has a decrease in the molof gas from reactants to products (as in a), the reaction will contract in volume ()V negative), and thesurroundings will do compression work on the system. When there is no change in the mol of gas fromreactants to products (as in c), )V = 0 and w = 0.

76. w = -P)V; )n = mol gaseous products - mol gaseous reactants. Only gases can do PV work (weignore solids and liquids). When a balanced reaction has more mol of product gases than mol ofreactant gases ()n positive), the reaction will expand in volume ()V positive) and the system will dowork on the surroundings. For example, in reaction c, )n = 2 - 0 = 2 mol, and this reaction would doexpansion work against the surroundings. When a balanced reaction has a decrease in the mol of gasfrom reactants to products ()n negative), the reaction will contract in volume ()V negative) and thesurroundings will do compression work on the system, e.g., reaction a where )n = 0 - 1 = -1. When there is no change in the mol of gas from reactants to products, )V = 0 and


w = 0, e.g., reaction b where )n = 2 - 2 = 0.

When )V > 0 ()n > 0), then w < 0 and system does work on the surroundings (c and e).

When )V < 0 ()n < 0), then w > 0 and the surroundings do work on the system (a and d).

When )V = 0 ()n = 0), then w = 0 (b).

overall step 1 step 277. )E = )E + )E ; This is a cyclic process which means that the overall initial state and

overall step 1 step 2final state are the same. Since )E is a state function, )E = 0 and )E = -)E .

step 1)E = q + w = 45 J + (-10. J) = 35 J

step 2 step 1)E = -)E = -35 J = q + w, -35 J = -60 J + w, w = 25 J

1 1 2 278. w = -P)V; We need the final volume of the gas. Since T and n are constant, P V = P V .

= 75.0 L

w = - P)V = -2.00 atm (75.0 L - 10.0 L) = -130. L atm × = -13.2 kJ = work

79. Heat loss by hot water = heat gain by cold water; Keeping all quantities positive to avoid sign errors:

hot × m × (55.0°C - 37.0°C) = × 90.0 g × (37.0°C - 22.0°C)

hotm = = 75.0 g hot water needed

2 280. 2 K(s) + 2 H O(l) ÷ 2 KOH(aq) + H (g), )H° = 2(-481 kJ) - 2(-286 kJ) = -390. kJ

5.00 g K × = -24.9 kJ of heat released upon reaction of 5.00 g of potassium.

24,900 J = × (1.00 × 10 g) × )T, )T = = 5.96°C3

Final temperature = 24.0 + 5.96 = 30.0°C

281. HCl(aq) + NaOH(aq) ÷ H O(l) + NaCl(aq) )H = -56 kJ

0.2000 L × = 8.00 × 10 mol HCl-2

0.1500 L × = 7.50 × 10 mol NaOH-2


Since the balanced reaction requires a 1:1 mol ratio between HCl and NaOH, and since fewer mol ofNaOH are actually present as compared to HCl, then NaOH is the limiting reagent.

Heat released = 7.50 × 10 mol NaOH × = -4.2 kJ heat released-2

82. The specific heat of water is 4.18 J/gC°C, which is equal to 4.18 kJ/kgC°C.

2We have 1.00 kg of H O, so: 1.00 kg × = 4.18 kJ/°C

2This is the portion of the heat capacity that can be attributed to H O.

cal cal Total heat capacity = C + , C = 10.84 - 4.18 = 6.66 kJ/°C

83. Heat released = 1.056 g × 26.42 kJ/g = 27.90 kJ = Heat gain by water and calorimeter

Heat gain = 27.90 kJ = × 0.987 kg × )T +

27.90 = (4.13 + 6.66) )T = 10.79 )T, )T = 2.586°C

f f2.586°C = T - 23.32°C, T = 25.91°C

84. To avoid fractions, let's first calculate )H for the reaction:

26 FeO(s) + 6 CO(g) ÷ 6 Fe(s) + 6 CO (g)

2 3 4 6 FeO + 2 CO ÷ 2 Fe O + 2 CO )H° = -2(18 kJ)

3 4 2 2 3 2 Fe O + CO ÷ 3 Fe O + CO )H° = -(-39 kJ)

2 3 2 3 Fe O + 9 CO ÷ 6 Fe + 9 CO )H° = 3(-23 kJ)

26 FeO(s) + 6 CO(g) ÷ 6 Fe(s) + 6 CO (g) )H° = -66 kJ

2So for: FeO(s) + CO(g) ÷ Fe(s) + CO (g) )H° = = -11 kJ

85. a. )H° = 3 mol (227 kJ/mol) - 1 mol (49 kJ/mol) = 632 kJ

2 2 6 6b. Since 3 C H (g) is higher in energy than C H (l), acetylene will release more energy per gramwhen burned in air.

286. Cu(s) + 1/2 O (g) ÷ CuO(s)


2 Cu(s) + CuO(s) ÷ Cu O(s) )H° = -(11 kJ)

2 2 Cu O(s) + 1/2 O (g) ÷ 2 CuO(s) )H° = -1/2(288 kJ)

2 Cu(s) + 1/2 O (g) ÷ CuO(s) )H° = -155 kJ =

2 4 3 3 287. a. C H (g) + O (g) ÷ CH CHO(g) + O (g), )H° = -166 kJ - [143 kJ + 52 kJ] = -361 kJ

3 2 2b. O (g) + NO(g) ÷ NO (g) + O (g), )H° = 34 kJ - [90. kJ + 143 kJ] = -199 kJ

3 2 2 4c. SO (g) + H O(l) ÷ H SO (aq), )H° = -909 kJ - [-396 kJ + (-286 kJ)] = -227 kJ

2 2d. 2 NO(g) + O (g) ÷ 2 NO (g), )H° = 2(34) kJ - 2(90.) kJ = -112 kJ

Challenge Problems

88. Only when there is a volume change can PV work be done. In pathway 1 (steps 1 + 2), only the firststep does PV work (step 2 has a constant volume of 30.0 L). In pathway 2 (steps 3 + 4), only step 4does PV work (step 3 has a constant volume of 10.0 L).

Pathway 1: w = -P)V = -2.00 atm (30.0 L - 10.0 L) = -40.0 L atm × = -4.05 × 10 J3

Pathway 2: w = -P)V = -1.00 atm (30.0 L - 10.0 L) = -20.0 L atm = -2.03 × 10 J3

Note: The sign is (-) because the system is doing work on the surroundings (an expansion).

We get a different value of work for the two pathways which both have the same initial and finalstates. Since w depends on the pathway, work cannot be a state function.

12 22 11 2 2 289. a. C H O (s) + 12 O (g) ÷ 12 CO (g) + 11 H O(l)

vb. A bomb calorimeter is at constant volume, so heat released = q = )E:

12 22 11)E = = -5630 kJ/mol C H O

c. Since PV = nRT, then P)V = RT)n where )n = mol gaseous products - mol gaseous reactants.

)H = )E + P)V = )E + RT)n

For this reaction, )n = 12 -12 = 0, so )H = )E = -5630 kJ/mol.

90. Energy needed = = 3.3 × 10 kJ/hr5

Energy from sun = 1.0 kW/m = 1000 W/m = 2 2


10,000 m × = 3.6 × 10 kJ/hr2 7

% efficiency = × 100 = × 100 = 0.92%

91. Energy used in 8.0 hours = 40. kWh = = 1.4 × 10 kJ5

Energy from the sun in 8.0 hours = × 8.0 h = 2.9 × 10 kJ/m4 2

Only 13% of the sunlight is converted into electricity:

0.13 × (2.9 × 10 kJ/m ) × Area = 1.4 × 10 kJ, Area = 37 m4 2 5 2

3 2 3 3 2 292. a. 2 HNO (aq) + Na CO (s) ÷ 2 NaNO (aq) + H O(l) + CO (g)

)H° = [2(-467 kJ) + (-286 kJ) + (-393.5 kJ)] - [2(-207 kJ) + (-1131 kJ)] = -69 kJ

2.0 × 10 gallons × = 1.1 × 10 g of concentrated nitric acid solution4 8

31.1 × 10 g solution × = 7.7 × 10 g HNO8 7

3 2 37.7 × 10 g HNO × = 6.5 × 10 g Na CO7 7

3There are (7.7 × 10 /63.02) mol of HNO from the previous calculation. There are 69 kJ of heat7

evolved for every two moles of nitric acid neutralized. Combining these two results:

37.7 × 10 g HNO × = -4.2 × 10 kJ7 7

b. They feared the heat generated by the neutralization reaction would vaporize the unreacted nitricacid, causing widespread airborne contamination.

93. 400 kcal × = 1.67 × 10 kJ . 2 × 10 kJ3 3

PE = mgz = = 160 J . 200 J

200 J of energy are needed to climb one step. The total number of steps to climb are:

2 × 10 J × = 1 × 10 steps6 4


2 2 294. H (g) +1/2 O (g) ÷ H O(1) )H = = -285.8 kJ

w = -P)V; Since PV = nRT, then at constant T and P, P)V = RT)n where )n = mol gaseous

products - mol gaseous reactants.

2 2 2For: 2 H O(1) ÷ 2 H (g) + O (g), )H = -2(-285.8 kJ) = 571.6 kJ and )n = 3 - 0 = 3

)H = )E + P)V = )E + RT)n, )E = )H - RT)n = 571.6 × 10 J - 8.3145 J/molCK (298 K)(3 mol)3

)E = 5.716 × 10 J - 7430 J = 5.642 × 10 J = 564.2 kJ5 5

chapter six thermochemistry · thermochemistry questions 9. a coff ee-cup calor imeter is at cons...

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